A `100 Omega` resistance and a capacitor of `100 Omega` reactance are connected in series across a `220` V source. When the capacitor is `50%` charged, the peak value of the displacement current is

Options

A. `4.4 A`
B. `11sqrt(2) A`
C. `2.2 A`
D. `11 A`

Explanation

To solve the problem, we need to find the peak value of the displacement current when a 100 Ω resistance and a capacitor with 100 Ω reactance are connected in series across a 220 V source.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = 100 Ω
- Capacitive Reactance (Xc) = 100 Ω
- Voltage (Vrms) = 220 V

2. **Calculate the Impedance (Z) of the Circuit:**
The impedance in an RC circuit is given by:
\[ Z = \sqrt{R^2 + X_c^2} \]
Substituting the values:
\[ Z = \sqrt{(100)^2 + (100)^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 100\sqrt{2} \, \Omega \]

3. **Convert the RMS Voltage to Peak Voltage (Vmax):**
The peak voltage (Vmax) can be calculated from the RMS voltage using the formula:
\[ V_{max} = V_{rms} \times \sqrt{2} \]
Substituting the given value:
\[ V_{max} = 220 \times \sqrt{2} \approx 220 \times 1.414 = 311.13 \, V \]

4. **Calculate the Peak Current (Imax):**
The peak current can be calculated using Ohm's law:
\[ I_{max} = \frac{V_{max}}{Z} \]
Substituting the values:
\[ I_{max} = \frac{311.13}{100\sqrt{2}} = \frac{311.13}{141.42} \approx 2.2 \, A \]

5. **Determine the Peak Value of the Displacement Current:**
The peak value of the displacement current (Id) is equal to the peak current (Imax) in the circuit since the displacement current in a capacitor is equal to the conduction current in the circuit:
\[ I_d = I_{max} = 2.2 \, A \]

### Final Answer:
The peak value of the displacement current is **2.2 A**.

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